## Online Course Discussion Forum

### Please help with winter camp Number theory p19-21

not sure about where to start. any help would be appreciated!

Update: I now need help on number theory blue p15, p19 and p20. thanks!

So far, I have only finished 15.

Hint: Prime Factorize 2013 and remember 7!/6! = 7 (how can you use this for 2013 similarly?)

Hint for after you prime factorize: you must MINIMIZE a_0 + b_0. 61 is prime, 59 as well. How can you build an extra "60" term in the denominator?

Number Theory 19: Note the problem is equivalent to "What is the smallest value of $k$ for which $k(k+1)(2k+1)$ is a multiple of $1200$. Thus, $3 \mid k(k+1)(2k+1)$, $2^4 \mid k(k+1)(2k+1)$, and $5^2 \mid k(k+1)(2k+1)$. Use this to narrow down possible choices for $k$.

Number Theory 20: Since $a = 2^n \cdot 3^m$, $a^6 = 2^{6n} \cdot 3^{6m}$, and $$\begin{aligned} 6^a &= (2 \cdot 3)^a \\ &= (2 \cdot 3)^{2^n \cdot 3^m} \\ &= 2^{2^n \cdot 3^m} \cdot 3^{2^n \cdot 3^m} \end{aligned}.$$ Look at the exponents of both. What do you need to ask of $n$ and $m$ so that $a^6 \mid 6^a$?

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